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All People in Canada are the Same Age
This "proof" will attempt to show that all people in Canada are the same
age, by showing by induction that the following statement (which we'll
call "S(n)" for short) is
true for all natural numbers n:
In any group of n people, everyone in that group has the same
The conclusion follows from that statement by letting n be the the
number of people in Canada.
If you're a little shaky on the principle of induction (which this proof
uses), there's a brief summary of it below.
The Fallacious Proof of Statement S(n):
Step 1: In any group
that consists of just one person, everybody in the group has
the same age, because after all there is only one person!
Step 2: Therefore, statement S(1) is true.
Step 3: The next stage in the induction argument is to prove that,
is true for one number (say n=k),
it is also true for the
next number (that is, n = k+1).
Step 4: We can do this by (1) assuming that,
in every group of k people,
everyone has the same age;
then (2) deducing from it that, in
every group of k+1 people, everyone has the same age.
Step 5: Let G be an arbitrary group of k+1 people;
we just need to show that every member of G has the same age.
Step 6: To do this, we just need to show that, if P and
Q are any members of G, then they have the same age.
Step 7: Consider everybody in G except
P. These people form a group of k people, so they must all have the
(since we are assuming that, in any group of k people, everyone has the
Step 8: Consider everybody in
G except Q. Again, they form a group of k people,
so they must all have the same age.
Step 9: Let R be someone else in G other than P or Q.
Step 10: Since Q and R each belong to
the group considered in step 7, they are the same age.
Step 11: Since P and R each belong to
the group considered in step 8, they are the same age.
Step 12: Since Q and R are the same age, and
P and R are the same age, it follows that P and Q
are the same age.
Step 13: We have now seen that, if we consider
any two people P and Q in G, they have the
same age. It follows that everyone in G has the same age.
Step 14: The proof is now complete: we have shown that
the statement is true for n=1, and we have shown that
whenever it is true for n=k it is also true for
n=k+1, so by induction it is true for all n.
See if you can figure out in which step the fallacy lies.
When you think you've figured it out, click on that step and
the computer will tell you whether you are correct or not, and
will give an additional explanation of why that step is or isn't valid.
See how many tries it takes you to correctly identify the fallacious step!
A Brief Review of the Principle of Induction
The principle of mathematical induction says this: Suppose you have
a set of natural numbers (natural numbers are the numbers
1, 2, 3, 4, . . . ). Suppose that 1 is in the set. Suppose also
that, whenever n is in the set, n+1 is also in the set. Then
every natural number is in the set.
To state it more informally: suppose you have the number 1 in your
collection, and for each number that you have in the collection, you
also have it plus 1 in your collection. Then you have all the natural
Intuitively, the idea is that if you start with the number 1, and
keep on adding 1 to it, you will eventually get to every number.
The principle of induction is extremely important because it allows
one to prove many results that are much more difficult to prove in
other ways. The most common application is when one has a statement
one wants to prove about each natural number. It may be quite difficult
to prove the statement directly, but easy to derive the truth of the
statement about n+1 from the truth of the statement about n. In that
case, one appeals to the principle of induction by showing
If you can prove those two things, then the principle of induction says
that the statement must be true for all natural numbers. (Reason:
let S be the set of numbers for which the statement is true. Item
1 says that 1 is in the set, and item 2 says that, whenever one number
n is in the set, n+1 is also in the set. Therefore, all numbers are in
- The statement is true when n=1.
- Whenever the statement is true for one number n, then
it's also true for the next number n+1.
As an example, consider proving that 1+2+3+· · ·+n = n(n+1)/2.
To try to prove that equality for a general,
unspecified n just by algebraic manipulations is
very difficult. But it's easy to prove by induction, because
it's true when n=1 (1 = 1(1+1)/2), and whenever it's true for
one number n, that means 1+2+3+· · ·+n = n(n+1)/2, so
1+2+3+· · ·+n+(n+1) = n(n+1)/2 + (n+1) = (n+1)(n+2)/2, so
it's also true for n+1. These two facts, combined with the
principle of induction, mean that it's true for all n.
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